$\begingroup$we have been told in arithmetic that the remainder of an integer divided by 5 is the same as the remainder of the division of the rightmost digit by 5 how to use modular properties to prove this?
$\endgroup$12 Answers
$\begingroup$Consider a number $\overline{a_1a_2a_3a_4....a_n}$. Now this can be written as $$ a_1\times 10^{n-1} +a_2\times 10^{n-2}+.....+a_{n-1}\times 10 +a_n$$ Now when you reduce this by $5$ you will get some remainder say $r$. We have $$ a_1\times 10^{n-1}n +a_2\times 10^{n-2}+.....+a_{n-1}\times 10 +a_n\equiv r\pmod5$$ $$\implies a_n\equiv r\pmod5$$.
$\endgroup$7$\begingroup$Hint: In decimal place-value notation, $edcba = 10\times edcb + a$. More generally, if $a$ is the right-most digit of a number $n$, then $n=10\times m + a$ for an appropriate number $m$.
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